# Some quick calc questions

don23
ok, first what is e^t times e^-t?
and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?
is it -costi + sintj +(t^2/2)k?

thank you!

ludi_srbin
don23 said:
ok, first what is e^t times e^-t?
and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?
is it -costi + sintj +(t^2/2)k?

thank you!

Well e^-t x e^t is same as e^t-t so It should be e^0=1.

I haven't started on Integrals yet.

bomba923
I believe

$$e^t e^{ - t} = e^0 = 1$$

and

$$\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2}$$
Just remember to take the limits of the integral.
You have antiderivative, now apply the algebra

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Homework Helper
$$\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2} + C$$
bomba923, I think you've made a mistake here,
First, I think the question reads:
$$\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt$$
And second, where does the little 'C' come from?
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don23 said:
is it -costi + sintj +(t^2/2)k?
Nope, it isn't.
In your post, you didn't state the variable of integration. So I assume it is dt.
So:
$$\int (\sin (it) + \cos (jt) + kt) dt = \int \sin (it) dt + \int \cos (jt) dt + \int kt dt$$
$$= \int \frac{1}{i} \sin (it) d(it) + \int \frac{1}{j} \cos (jt) d(jt) + k \int t dt$$
$$= \frac{1}{i} \int \sin (it) d(it) + \frac{1}{j} \int \cos (jt) d(jt) + k \int t dt = ...$$
Can you go from here?
Viet Dao,

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Buzzer
VietDao29 said:
First, I think the question reads:
$$\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt$$

You have confused me here.

$$\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt$$

$$i + j + \frac{{\pi ^2 }}{2}k$$

Homework Helper
Buzzer said:
You have confused me here.

$$\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt$$

If I am asking some question like that, I am going to put the i, and j in front of the cos and sin functions.
Something like this:
i sint + j cost + kt.

Buzzer said:
$$i + j + \frac{{\pi ^2 }}{2}k$$

And even if the question reads like this:
$$\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt$$
That is still not the correct answer.
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EDIT:
If the question happens to be:
int (i sint + j cost + kt) dt
You can use bomba923's suggestion. You should omit the constant of integration 'C' in his answer.
Viet Dao,

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Buzzer
VietDao29 said:
If I am asking some question like that, I am going to put the i, and j in front of the cos and sin functions.
Something like this:
i sint + j cost + kt.
Ok, but I learned to put i, j and k at the end when I did vectors...

I did it again and got the same answer. Where does 2i come from?

EDIT

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Homework Helper
Certainly anyone taking calculus should have learned long ago that axbx= ax+y. In particular, axa-x= a0= 1. The fact that a= e is irrelevant.

Yes, the original post's "integral from 0 to pi of (sinti+costj+tk)" should have been written
$$\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k} dt[/itex] to distinguish it form [tex]\int_0^{\pi}{sin ti+ cos tj+ tk} dt[/itex] but I think the "i, j, k" suggested vectors rather than numbers. don23, It is true that [tex]\int {(sin t)i+ (cos t)j+ (t)k}dt= -cos(t)i+ sin(t)j+ (\frac{1}{2}t^2+ C$$
but you forgot to evaluate it between 0 and $\pi$.

$$(-cos \pi)i+ (sin /pi)j+ \frac{1}{2}(/pi)^2= i+ 0j+ \frac{\pi^2}{2}k$$
and
$$(-cos 0)i+ (sin 0)i+ \frac{1}{2}(0)^2= -i+ 0j+ 0k$$
so
$$\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k}dt= 2i +\frac{\pi^2}{2}k$$

bomba923
Thank you Halls of Ivy
(you've vindicated me!)